Is there any number $n$ such that $nm=0$, $n\neq 0$, and $m\neq 0$?

Question

I answered a question about whether zero is prime or composite on Khan Academy a while ago. Since then, two people have commented on my answer, asking another question that I don't know the answer to.

Are there zero divisors (numbers such that, when multiplied by some nonzero number, the product is zero) other than zero? If so, what are they? Wouldn't that go against the Zero Product Property, which states that if you take two numbers, $n$ and $m$, and multiply them together to get zero, then either $n$ or $m$ is equal to zero? I.e. if $nm=0$, then either $n=0$ and $m=0$.

$\mathbb{Z} / 6\mathbb{Z}$ is the multiplicative group of integers modulo $6$. Read here for reference: http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n — Yiyuan Lee, Mar 25 '14 at 13:30
@YiyuanLee: It is not a group but merely a monoid (there are no zero divisors in a group since they do not have an inverse). The group $(ℤ/6ℤ)^*$ exists, but it does not contain the elements 2 and 3. — Ansgar Esztermann, Mar 25 '14 at 13:58
Note $\ $ Following the link reveals the context is "6th grade math (US)". So mention of fields and domains is probably over the head of the audience. Maybe they could understand zero-divisors in modular integer arithmetic if suitably explained. — Bill Dubuque, Mar 25 '14 at 15:42

fgp · Accepted Answer · 2014-03-25T15:30:38.600

4

In fields (e.g., $\mathbb{Q}$,$\mathbb{R}$,$\mathbb{C}$) and integral domains (e.g. $\mathbb{Z},\mathbb{Z}[x]$), no such pair of "numbers" $n,m$ can exist.

For fields, this follows from the invertibility of multiplication. If $nm = 0$ but $n \neq 0$, then $m = n^{-1}0 = 0$, and vice versa.

For integral domains, this is an axiom - and it's precisely what differentiates integral domains from rings.

In arbitrary rings, such numbers can exists. E.g., in the ring $\mathbb{Z_4} = \{0,1,2,3\}$ you have $2\cdot 2 = 4 = 0$.

edited Mar 25 '14 at 15:30

answered Mar 25 '14 at 13:53

fgp

21,050

Better to use integral domain (domain of integrity is very old-fashioned and little-used nowadays). – Bill Dubuque Mar 25 '14 at 15:29
@BillDubuque Fixed. I keep making that mistake - probably because the german word is "Integritätsbereich", which literally translates to "integrity domain". Oh well... – fgp Mar 25 '14 at 15:32
@fgp: Is it really valid to say 4 = 0 directly? Is there not something more to it. – Nick Mar 25 '14 at 15:33
1

See also this question on the etymology of "integral domain". – Bill Dubuque Mar 25 '14 at 15:36
3

@Nick It's absolutely valid, and there's really nothing more to it; in $\mathbb{Z}_/4\mathbb{Z}$, $4$ and $0$ are different names for the same element. – Steven Stadnicki Mar 25 '14 at 15:36

score 3 · Answer 2 · answered Mar 25 '14 at 13:12

3

That depends on what you mean by a "number". Are we talking about real numbers? Complex numbers? Split-complex? Quaternions? Octonions? Arbitrary rings?

Certainly, if we state that $n$ and $m$ are real numbers, then this statement must hold.

answered Mar 25 '14 at 13:12

Ben Grossmann

225,327

Is there any number $n$ such that $nm=0$, $n\neq 0$, and $m\neq 0$?

2 Answers2